Solved Examples Based On Determination Of Forces In All The Members Of The Truss II

Example: Find the forces in the members of the truss shown in figure.

Solution:

Step 1: Determine of inclination

tan θ = 3/4 ⇒θ tan–1 (3/4) = 36.86o

⇒ sin θ = 0.6 and cos = 0.8

Step 2: Considering the free body diagram of joint B

ΣFx = 0 ⇒ FBC – FBD cos θ = 0 …(1)

ΣFy = 0 ⇒ FBD sin θ – 1000 = 0 …(2)

FBD = 1000 / sin θ = 1000 / 0.6 = 1666.67 N (Compressive)

From eq. (1)

FBC – FBD cos θ = 0

FBC = FBD cos θ = 1666.67 × 0.8

FBC = 1333.34 N (Tension)

Step 3: Consider the free body diagram of joint C

ΣFy = 0 ⇒ 1000 – FCD = 0 …(3)

FCD = 1000 N (Compressive)

ΣFx = 0 ⇒ FCA = FCB …(4)

FCA = 1333.43 N (Tensile)

Step 4 Consider the free body diagram of joint D

ΣFy = 0 ⇒ FDE sin θ + FBA sin θ – FDC – FBD sin θ = 0

0.6 FDE + 0.6 FDA – 1000 – 1666.67 × 0.6 = 0

FDE + FDA = 333.34 …(5)

ΣFx = 0 ⇒ FDA cos θ – FDE cos θ + FDB cos θ = 0

FDA – FDE = –FDB = –1666.67 …(6)

On solving (5) and (6)

FDA = 833.33 N (Tensile)

FDE = 3333.33 – FDA

= 3333.33 – 833.33

= 2500 N (Compressive)

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